Your LED is getting power from the Microcontroller but, you have both pins connected to the Microcontroller. If you plan to do it like this (and you shouldn't - but if you did), you would need to set the Microcontroller pin connected to the a note (+) pin of ?LED as an output pin and set it high (1). Then set the Microcontroller pin connected to LeD cathode pin as an output and set it low (0).

The trouble with that arrangement is that nothing limits the amount of current flowing through the Led. Assuming your Microcontroller is running at 5v, you need to add a resistor (150 to 330 ohm) in series with the LED and one of the micro pins.

If you were able to set your pins conneded to the LED as described in the first paragraph, you may have damaged one or both (and possibly the LEd).

Test that the LED works by connecting it to your power (but using a resistor in series to control current).

Then try again with two different pins. - you do have a programmer, compiler and all that is needed to get your program to your micro, correct?

There may be other issues but the first one I see is the capacitor at the far left in the 1st photo, near the regulator. It has no contact with the regulator, as connections do not carry across the central "trench".

The ground wire is also not connected to the regulator, see 3rd photo. That diode on the power rails isn't doing anything either. The power wire to the µP goes nowhere.

I suspect you need to study which breadboard pins are interconnected and which are not.

Also, your battery (-) should be connected to the ground strip (black or dark blue) and battery (+) should be connected to the regulator's input. The regulator's output should connect to the board red power strips to give you 5 volts on this strips.

As connected, you have battery voltage at the power strips and the regulator is doing nothing.

Hi Vead. As a tip, most newer cellphone chargers give out around 5 volt. If you have some that is not used lying around. It may be used as power source for bread board. As those 9 volts battery wear out quite fast.
At the board connection point use the 1000uF and 22nF in parallel with the voltage source. You must of course have a voltmeter to check the polarity and that the voltage is correct. If it is somewhat lower than 5 volt, say 4.5 volt it does not have much to say

I have attached Image I want to make this type of power supply circuit

Q) in my previous post . i have uploaded three Images. Is it correct ? or I need to changes in connection

Click to expand...

Your breadboard is looking better near the regulator but...
- I don't see where you are connecting your battery. Battery Positive should connect to pin 1 (left pin when you can read the 7805 part number), battery (-) should connect to black power rail.

- your diode is connected to two points on the positive power rail - it is doing nothing.

- I am not sure you need the small yellow and black jumper wires in the middle of your power rail board. Use an ohm meter to check if your power rails are split at the mid-point of your board.

I have attached Image I want to make this type of power supply circuit..

Click to expand...

You need to add a diode in reverse orientation to bypass the regulator in the event the power supply is removed. The remaining charge on the output capacitor, or perhaps from the load, will otherwise hold the output voltage higher than the input voltage on the regulator, and this can damage it.

I would like to suggest that you start with a simpler breadboard project. A 555 based flasher for example. This will help you get familiar with how the breadboard works without all the complications of programming a microcontroller. A 555 can also run directly from the 9V so no need for a regulator.

No, I mean you need to add a path for the output capacitor to discharge to the input side of the regulator, so that it doesn't drive current back into the regulator. Read the data sheets for the regulator - it will show this in the example circuits.

No, he means one more diode - in reverse of normal current flow like this (D1). It is there "just in case". Note, the stripped side of D1 should be connected to pin1 of the 7805 and the other end of D1 should be connected to pin 3 o 7805.